Can we impose a boundary condition on the derivative of the wavefunction through the physical assumptions?

Question

Consider the Schrödinger equation for a particle in one dimension, where we have at least one boundary in the system (say the boundary is at $x=0$ and we are solving for $x>0$). Sometimes we want to impose a boundary condition in which the wavefunction vanishes (Dirichlet boundary condition).

We can indirectly impose this boundary condition through the physical assumptions by using an infinite potential outside the relevant region (like in the "particle in a box" model): $$ V(x<0)=\infty ~~~~\Longrightarrow ~~~~\psi(x=0)=0 $$ What if we want to impose a boundary condition in which the derivative of the wavefunction vanishes (Neumann boundary condition)? $$ ? ~~~~\Longrightarrow ~~~~ \left. \frac{\partial \psi}{\partial x} \right|_{x=0} = 0 $$ Is there a way to choose the potential, or maybe change something else in the Hamiltonian, in order to indirectly impose this boundary condition?

P.S. This question is not of great practical importance, it is more of a curiosity.

Answer 1

By mirroring $V(x)$ about $x = 0$, i.e., by setting $V(-x) = V(x)$, the wavefunction can be taken to be even or odd. The even solution satisfies the Neumann boundary condition since the derivative of an even function is odd and thus zero at $x = 0$.

Answer 2

It's not really a physical condition, but when one is doing R-matrix theory for scattering (which is arguably not for the faint of heart) the condition does come up. One resource I saw recently is a lecture by Hugo van der Hart (go to the slide titled Basic Applications).